Doepfer

I've noticed recently that the A-180 Multiple really cuts the voltage 
signal down quite substantially. I just got an MFB Dual ADSR and noticed 
that when i mult the Gate Output of my SH-101 by 3x, it drops the gate 
voltage down to where the MFB won't trigger anymore. Basically, i had it 
set up where the gate out went into the A-180, then the first 2 outs 
went to the two Gate Ins on the MFB and the 3rd out went to my A-140 EG. 
If i pulled the cable out that went to the A-140, the MFB would fire, 
but not with it inserted. However, it works if i use my 303, which has a 
hotter Gate signal. I checked the gate out voltages coming of the 101 
direct, then at each output of the A-180. I got the following:

Direct Out from 101: 11.56v
Into the A-180 w/no other cables inserted: same as above
with 2 cables inserted: 7v
with 3 cables inserted: 5.5v

This is quite a bit of voltage drop if you ask me. Is there anyway to 
mod the A-180 so it doesn't eat up all my voltages?

-Larry

It's strange because in theory the voltage should not drop since these are
parallel circuits-- only the current should change.

On Sun, Nov 15, 2009 at 2:16 PM, Computer Controlled <acidted@gmail.com>wrote:

>
>
> I've noticed recently that the A-180 Multiple really cuts the voltage
> signal down quite substantially. I just got an MFB Dual ADSR and noticed
> that when i mult the Gate Output of my SH-101 by 3x, it drops the gate
> voltage down to where the MFB won't trigger anymore. Basically, i had it
> set up where the gate out went into the A-180, then the first 2 outs
> went to the two Gate Ins on the MFB and the 3rd out went to my A-140 EG.
> If i pulled the cable out that went to the A-140, the MFB would fire,
> but not with it inserted. However, it works if i use my 303, which has a
> hotter Gate signal. I checked the gate out voltages coming of the 101
> direct, then at each output of the A-180. I got the following:
>
> Direct Out from 101: 11.56v
> Into the A-180 w/no other cables inserted: same as above
> with 2 cables inserted: 7v
> with 3 cables inserted: 5.5v
>
> This is quite a bit of voltage drop if you ask me. Is there anyway to
> mod the A-180 so it doesn't eat up all my voltages?
>
> -Larry
>  
>


[Non-text portions of this message have been removed]

That's why there are Buffered Multiples, passive ones drop the voltage 
levels. So they need to be buffered to keep them at original levels.

I just didn't think that the level would drop by so much!



Monroe Eskew wrote:

> It's strange because in theory the voltage should not drop since these are
> parallel circuits-- only the current should change.
>
> On Sun, Nov 15, 2009 at 2:16 PM, Computer Controlled <acidted@gmail.com>wrote:
>
>   
>> I've noticed recently that the A-180 Multiple really cuts the voltage
>> signal down quite substantially. I just got an MFB Dual ADSR and noticed
>> that when i mult the Gate Output of my SH-101 by 3x, it drops the gate
>> voltage down to where the MFB won't trigger anymore. Basically, i had it
>> set up where the gate out went into the A-180, then the first 2 outs
>> went to the two Gate Ins on the MFB and the 3rd out went to my A-140 EG.
>> If i pulled the cable out that went to the A-140, the MFB would fire,
>> but not with it inserted. However, it works if i use my 303, which has a
>> hotter Gate signal. I checked the gate out voltages coming of the 101
>> direct, then at each output of the A-180. I got the following:
>>
>> Direct Out from 101: 11.56v
>> Into the A-180 w/no other cables inserted: same as above
>> with 2 cables inserted: 7v
>> with 3 cables inserted: 5.5v
>>
>> This is quite a bit of voltage drop if you ask me. Is there anyway to
>> mod the A-180 so it doesn't eat up all my voltages?
>>
>> -Larry
>>  
>>
>>     
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>
>
>

Hi Guys

There seems to be some confusion about what's going on with loading, volts, current etc.  The good news is that Ohm's Law isn't about to be disproved anytime soon. Volts and current are linked in Ohm's law by resistance, in this case the output resistance of the SH101 - this is about 22k and it connects to +15V in the SH101 gate out circuit. 

The MFB EG acts as a load and two in parallel act as a greater load, i.e. more current will flow.  If for argument's sake the MFB has an input resistance of 10k then two EGs in parallel would then give a load of 5k. I'll ignore the A140 since that that has a very high input resistance (mine are around 5M to 7M). 

The arithmetic for this example is that 15V is loaded by 22k out plus a 5k load = 27k. This acts as a potential divider thus 15/27 x 5 = 2.8V. Not very big! So the more current out the more the voltage drop across the output resistance and hence a smaller gate voltage. So yes it needs a buffer to isolate the SH101 output's high output resistance; I use an A185-1. If you just had A140s then these have such a high resistance that it shouldn't make any difference.

Hope that helps clear up the mystery.
I'd be inteested if anyone can tell me what the MFB EG input resistance actually is.

Cheers
Tony



--- In Doepfer_a100@yahoogroups.com, Monroe Eskew <monroe.eskew@...> wrote:

>
> Pardon if this is a dumb question, but are you sure you're measuring voltage
> and not power?  Is Ohm's law really that blatantly false?  I've noticed VCOs
> don't respond to voltage only; they are influenced by the amperage as well.
> 
> On Sun, Nov 15, 2009 at 2:43 PM, Computer Controlled <acidted@...>wrote:
> 
> >
> >
> > That's why there are Buffered Multiples, passive ones drop the voltage
> > levels. So they need to be buffered to keep them at original levels.
> >
> > I just didn't think that the level would drop by so much!
> >
> >
> > Monroe Eskew wrote:
> > > It's strange because in theory the voltage should not drop since these
> > are
> > > parallel circuits-- only the current should change.
> > >
> > > On Sun, Nov 15, 2009 at 2:16 PM, Computer Controlled <acidted@...<acidted%40gmail.com>
> > >wrote:
> > >
> > >
> > >> I've noticed recently that the A-180 Multiple really cuts the voltage
> > >> signal down quite substantially. I just got an MFB Dual ADSR and noticed
> > >> that when i mult the Gate Output of my SH-101 by 3x, it drops the gate
> > >> voltage down to where the MFB won't trigger anymore. Basically, i had it
> > >> set up where the gate out went into the A-180, then the first 2 outs
> > >> went to the two Gate Ins on the MFB and the 3rd out went to my A-140 EG.
> > >> If i pulled the cable out that went to the A-140, the MFB would fire,
> > >> but not with it inserted. However, it works if i use my 303, which has a
> > >> hotter Gate signal. I checked the gate out voltages coming of the 101
> > >> direct, then at each output of the A-180. I got the following:
> > >>
> > >> Direct Out from 101: 11.56v
> > >> Into the A-180 w/no other cables inserted: same as above
> > >> with 2 cables inserted: 7v
> > >> with 3 cables inserted: 5.5v
> > >>
> > >> This is quite a bit of voltage drop if you ask me. Is there anyway to
> > >> mod the A-180 so it doesn't eat up all my voltages?
> > >>
> > >> -Larry
> > >>
> > >>
> > >>
> > >
> > >
> > > [Non-text portions of this message have been removed]
> > >
> > >
> > >
> > > ------------------------------------
> > >
> > > Yahoo! Groups Links
> > >
> > >
> > >
> > >
> > >
> >  
> >
> 
> 
> [Non-text portions of this message have been removed]
>

Pardon if this is a dumb question, but are you sure you're measuring voltage
and not power?  Is Ohm's law really that blatantly false?  I've noticed VCOs
don't respond to voltage only; they are influenced by the amperage as well.

On Sun, Nov 15, 2009 at 2:43 PM, Computer Controlled <acidted@gmail.com>wrote:

>
>
> That's why there are Buffered Multiples, passive ones drop the voltage
> levels. So they need to be buffered to keep them at original levels.
>
> I just didn't think that the level would drop by so much!
>
>
> Monroe Eskew wrote:
> > It's strange because in theory the voltage should not drop since these
> are
> > parallel circuits-- only the current should change.
> >
> > On Sun, Nov 15, 2009 at 2:16 PM, Computer Controlled <acidted@gmail.com<acidted%40gmail.com>
> >wrote:
> >
> >
> >> I've noticed recently that the A-180 Multiple really cuts the voltage
> >> signal down quite substantially. I just got an MFB Dual ADSR and noticed
> >> that when i mult the Gate Output of my SH-101 by 3x, it drops the gate
> >> voltage down to where the MFB won't trigger anymore. Basically, i had it
> >> set up where the gate out went into the A-180, then the first 2 outs
> >> went to the two Gate Ins on the MFB and the 3rd out went to my A-140 EG.
> >> If i pulled the cable out that went to the A-140, the MFB would fire,
> >> but not with it inserted. However, it works if i use my 303, which has a
> >> hotter Gate signal. I checked the gate out voltages coming of the 101
> >> direct, then at each output of the A-180. I got the following:
> >>
> >> Direct Out from 101: 11.56v
> >> Into the A-180 w/no other cables inserted: same as above
> >> with 2 cables inserted: 7v
> >> with 3 cables inserted: 5.5v
> >>
> >> This is quite a bit of voltage drop if you ask me. Is there anyway to
> >> mod the A-180 so it doesn't eat up all my voltages?
> >>
> >> -Larry
> >>
> >>
> >>
> >
> >
> > [Non-text portions of this message have been removed]
> >
> >
> >
> > ------------------------------------
> >
> > Yahoo! Groups Links
> >
> >
> >
> >
> >
>  
>


[Non-text portions of this message have been removed]

Ah, sorry, I didn't really read the details.  Sounds like he should be able
to fix it by soldering high value resistors to the inputs of the MFB EG... I
think.

On Sun, Nov 15, 2009 at 3:49 PM, Synthomaniac <tony.steventon@talk21.com>wrote:

>
>
> Hi Guys
>
> There seems to be some confusion about what's going on with loading, volts,
> current etc. The good news is that Ohm's Law isn't about to be disproved
> anytime soon. Volts and current are linked in Ohm's law by resistance, in
> this case the output resistance of the SH101 - this is about 22k and it
> connects to +15V in the SH101 gate out circuit.
>
> The MFB EG acts as a load and two in parallel act as a greater load, i.e.
> more current will flow. If for argument's sake the MFB has an input
> resistance of 10k then two EGs in parallel would then give a load of 5k.
> I'll ignore the A140 since that that has a very high input resistance (mine
> are around 5M to 7M).
>
> The arithmetic for this example is that 15V is loaded by 22k out plus a 5k
> load = 27k. This acts as a potential divider thus 15/27 x 5 = 2.8V. Not very
> big! So the more current out the more the voltage drop across the output
> resistance and hence a smaller gate voltage. So yes it needs a buffer to
> isolate the SH101 output's high output resistance; I use an A185-1. If you
> just had A140s then these have such a high resistance that it shouldn't make
> any difference.
>
> Hope that helps clear up the mystery.
> I'd be inteested if anyone can tell me what the MFB EG input resistance
> actually is.
>
> Cheers
> Tony
>
>
> --- In Doepfer_a100@yahoogroups.com <Doepfer_a100%40yahoogroups.com>,
> Monroe Eskew <monroe.eskew@...> wrote:
> >
> > Pardon if this is a dumb question, but are you sure you're measuring
> voltage
> > and not power? Is Ohm's law really that blatantly false? I've noticed
> VCOs
> > don't respond to voltage only; they are influenced by the amperage as
> well.
> >
> > On Sun, Nov 15, 2009 at 2:43 PM, Computer Controlled <acidted@...>wrote:
>
> >
> > >
> > >
> > > That's why there are Buffered Multiples, passive ones drop the voltage
> > > levels. So they need to be buffered to keep them at original levels.
> > >
> > > I just didn't think that the level would drop by so much!
> > >
> > >
> > > Monroe Eskew wrote:
> > > > It's strange because in theory the voltage should not drop since
> these
> > > are
> > > > parallel circuits-- only the current should change.
> > > >
> > > > On Sun, Nov 15, 2009 at 2:16 PM, Computer Controlled <acidted@
> ...<acidted%40gmail.com>
>
> > > >wrote:
> > > >
> > > >
> > > >> I've noticed recently that the A-180 Multiple really cuts the
> voltage
> > > >> signal down quite substantially. I just got an MFB Dual ADSR and
> noticed
> > > >> that when i mult the Gate Output of my SH-101 by 3x, it drops the
> gate
> > > >> voltage down to where the MFB won't trigger anymore. Basically, i
> had it
> > > >> set up where the gate out went into the A-180, then the first 2 outs
> > > >> went to the two Gate Ins on the MFB and the 3rd out went to my A-140
> EG.
> > > >> If i pulled the cable out that went to the A-140, the MFB would
> fire,
> > > >> but not with it inserted. However, it works if i use my 303, which
> has a
> > > >> hotter Gate signal. I checked the gate out voltages coming of the
> 101
> > > >> direct, then at each output of the A-180. I got the following:
> > > >>
> > > >> Direct Out from 101: 11.56v
> > > >> Into the A-180 w/no other cables inserted: same as above
> > > >> with 2 cables inserted: 7v
> > > >> with 3 cables inserted: 5.5v
> > > >>
> > > >> This is quite a bit of voltage drop if you ask me. Is there anyway
> to
> > > >> mod the A-180 so it doesn't eat up all my voltages?
> > > >>
> > > >> -Larry
> > > >>
> > > >>
> > > >>
> > > >
> > > >
> > > > [Non-text portions of this message have been removed]
> > > >
> > > >
> > > >
> > > > ------------------------------------
> > > >
> > > > Yahoo! Groups Links
> > > >
> > > >
> > > >
> > > >
> > > >
> > >
> > >
> >
> >
> > [Non-text portions of this message have been removed]
> >
>
>  
>


[Non-text portions of this message have been removed]

Hi Monroe,

Sorry to be the bearer of bad news, but soldering resistors to the inputs whilst reducing load current will only act as a further potential divider causing an even worse situation because it adds to the output resistance of the SH101.  The only way forward is to have an active buffer (e.g. A185-1 or Dieter may be able to suggest other alternatives). Most Doepfer modules have a low output resistance, normally 1k or lower which as you can see is significantly lower than the SH101. 

I must still stress that I don't actually know what the input resistance of the MFB EGs is and that neither has a fault. My gut feel is that the MFBs should be higher than the 10k that I suggested in my example. It would be useful if all manufacturers produce a spec sheet stating input and output resistances and voltage levels etc. It would help in determining compatability between different manufacturer's modules. I'll ping off a question to MFB to find out.

Cheers
Tony



--- In Doepfer_a100@yahoogroups.com, Monroe Eskew <monroe.eskew@...> wrote:

>
> Ah, sorry, I didn't really read the details.  Sounds like he should be able
> to fix it by soldering high value resistors to the inputs of the MFB EG... I
> think.
> 
> On Sun, Nov 15, 2009 at 3:49 PM, Synthomaniac <tony.steventon@...>wrote:
> 
> >
> >
> > Hi Guys
> >
> > There seems to be some confusion about what's going on with loading, volts,
> > current etc. The good news is that Ohm's Law isn't about to be disproved
> > anytime soon. Volts and current are linked in Ohm's law by resistance, in
> > this case the output resistance of the SH101 - this is about 22k and it
> > connects to +15V in the SH101 gate out circuit.
> >
> > The MFB EG acts as a load and two in parallel act as a greater load, i.e.
> > more current will flow. If for argument's sake the MFB has an input
> > resistance of 10k then two EGs in parallel would then give a load of 5k.
> > I'll ignore the A140 since that that has a very high input resistance (mine
> > are around 5M to 7M).
> >
> > The arithmetic for this example is that 15V is loaded by 22k out plus a 5k
> > load = 27k. This acts as a potential divider thus 15/27 x 5 = 2.8V. Not very
> > big! So the more current out the more the voltage drop across the output
> > resistance and hence a smaller gate voltage. So yes it needs a buffer to
> > isolate the SH101 output's high output resistance; I use an A185-1. If you
> > just had A140s then these have such a high resistance that it shouldn't make
> > any difference.
> >
> > Hope that helps clear up the mystery.
> > I'd be inteested if anyone can tell me what the MFB EG input resistance
> > actually is.
> >
> > Cheers
> > Tony
> >
> >
> > --- In Doepfer_a100@yahoogroups.com <Doepfer_a100%40yahoogroups.com>,
> > Monroe Eskew <monroe.eskew@> wrote:
> > >
> > > Pardon if this is a dumb question, but are you sure you're measuring
> > voltage
> > > and not power? Is Ohm's law really that blatantly false? I've noticed
> > VCOs
> > > don't respond to voltage only; they are influenced by the amperage as
> > well.
> > >
> > > On Sun, Nov 15, 2009 at 2:43 PM, Computer Controlled <acidted@>wrote:
> >
> > >
> > > >
> > > >
> > > > That's why there are Buffered Multiples, passive ones drop the voltage
> > > > levels. So they need to be buffered to keep them at original levels.
> > > >
> > > > I just didn't think that the level would drop by so much!
> > > >
> > > >
> > > > Monroe Eskew wrote:
> > > > > It's strange because in theory the voltage should not drop since
> > these
> > > > are
> > > > > parallel circuits-- only the current should change.
> > > > >
> > > > > On Sun, Nov 15, 2009 at 2:16 PM, Computer Controlled <acidted@
> > ...<acidted%40gmail.com>
> >
> > > > >wrote:
> > > > >
> > > > >
> > > > >> I've noticed recently that the A-180 Multiple really cuts the
> > voltage
> > > > >> signal down quite substantially. I just got an MFB Dual ADSR and
> > noticed
> > > > >> that when i mult the Gate Output of my SH-101 by 3x, it drops the
> > gate
> > > > >> voltage down to where the MFB won't trigger anymore. Basically, i
> > had it
> > > > >> set up where the gate out went into the A-180, then the first 2 outs
> > > > >> went to the two Gate Ins on the MFB and the 3rd out went to my A-140
> > EG.
> > > > >> If i pulled the cable out that went to the A-140, the MFB would
> > fire,
> > > > >> but not with it inserted. However, it works if i use my 303, which
> > has a
> > > > >> hotter Gate signal. I checked the gate out voltages coming of the
> > 101
> > > > >> direct, then at each output of the A-180. I got the following:
> > > > >>
> > > > >> Direct Out from 101: 11.56v
> > > > >> Into the A-180 w/no other cables inserted: same as above
> > > > >> with 2 cables inserted: 7v
> > > > >> with 3 cables inserted: 5.5v
> > > > >>
> > > > >> This is quite a bit of voltage drop if you ask me. Is there anyway
> > to
> > > > >> mod the A-180 so it doesn't eat up all my voltages?
> > > > >>
> > > > >> -Larry
> > > > >>
> > > > >>
> > > > >>
> > > > >
> > > > >
> > > > > [Non-text portions of this message have been removed]
> > > > >
> > > > >
> > > > >
> > > > > ------------------------------------
> > > > >
> > > > > Yahoo! Groups Links
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > >
> > > >
> > >
> > >
> > > [Non-text portions of this message have been removed]
> > >
> >
> >  
> >
> 
> 
> [Non-text portions of this message have been removed]
>

Hi Larry

You may try to change the values of the buffering stage in the SH-101. 
It is TR9 with R47 and R49 (close to the top left corner of the CPU).

You may reduce R47 to 4k7 and R49 to 22k.

Florian

> I've noticed recently that the A-180 Multiple really cuts the voltage
> signal down quite substantially. I just got an MFB Dual ADSR and noticed
> that when i mult the Gate Output of my SH-101 by 3x, it drops the gate
> voltage down to where the MFB won't trigger anymore. Basically, i had it
> set up where the gate out went into the A-180, then the first 2 outs
> went to the two Gate Ins on the MFB and the 3rd out went to my A-140 EG.
> If i pulled the cable out that went to the A-140, the MFB would fire,
> but not with it inserted. However, it works if i use my 303, which has a
> hotter Gate signal. I checked the gate out voltages coming of the 101
> direct, then at each output of the A-180. I got the following:
>
> Direct Out from 101: 11.56v
> Into the A-180 w/no other cables inserted: same as above
> with 2 cables inserted: 7v
> with 3 cables inserted: 5.5v
>
> This is quite a bit of voltage drop if you ask me. Is there anyway to
> mod the A-180 so it doesn't eat up all my voltages?
>
> -Larry

Some basic information about this issue:

The A-180 is a passive module - nothing but 2x4 connected jack sockets. It
works as if one would simply connect the jack plugs manually of if one would
use an Y-cable. Consequently the reason for the behaviour described is
actually not the A-180 but the inputs and outputs of the modules. Ideally
the output of each module should be able to drive any load without any
losses (technically spoken: zero ohm output impedance or resistance). And
ideally the input of each module should not cause any load (technically
spoken: infinite ohm output impedance or resistance). But in reality the
situation is different: each output has a limited driving capability (e.g. 1
k output resistance) and each input causes a load (e.g. 100k for most VCO CV
inputs). As a rule of thumb the relation between input and output resistance
causes the voltage loss: if a 1k output drives a  100k input one has about
1/100 of voltage drop. Not much for a gate voltage but already too much for
a VCO control input(that's why special buffers like A-185-1/2 are
recommended for VCO control). If more than one input is connected to an
output (e.g. by means of an A-180) the resulting load becomes smaller: as a
rule of thumb the load is divided by two if both inputs have the same load.
Otherwise it's a bit more difficult to calculate. For example: the A-140 has
a 10k load at it's input. If two A-140 inputs are connected together the
resulting load is 5k. If the resistance of the output that drives the two
A-140 is also in the 5k range the output voltage is divided by two. If it
was 12V without load it becomes ~ 6V with two A-140.

In the A-100 we payed attention that (with some exceptions) each source is
able to drive a customary number of units (e.g. 3-4 A-140 from a gate
source). The output resistance of A-100 modules is typ. 1k and the input
resistance 10k or more (e.g. 100k for VCO control inputs). Consequently
within the A-100 no problems with gate signals should occur. For voltages
that drive VCOs buffer modules (A-185-1/2) are recommended. These have a
much lower output impedance. If an external voltage source is used the
situation may be different. One needs to know the output impedance to
calculate the expected voltage drop. To be on the safe side a buffer module
can be used (even for gate signals).

Best wishes
Dieter Doepfer

yes, that's why when i use the A100 with other modular devices i, very often, use buffer modules even for gates clocking various other sequencers, etc...


Bakis Sirros - Parallel Worlds / Interconnected / Memory Geist

[Doepfer_a100] group owner

www. parallel - worlds - music. com

www. myspace. com/ parallelworldsmusic

www. myspace. com/ interconnectedmusic

www. myspace. com/ memorygeist

www. DiN. org. uk

www. musicamaximamagnetica. com

www. shimarecords. co. uk

www. rubberrecords. gr

Athens - Greece

--- On Mon, 11/16/09, yahoo@doepfer.de <yahoo@doepfer.de> wrote:

From: yahoo@doepfer.de <yahoo@doepfer.de>
Subject: AW: [Doepfer_a100] A-180
To: Doepfer_a100@yahoogroups.com
Date: Monday, November 16, 2009, 11:58 AM







 



  


    
      
      
      > I've noticed recently that the A-180 Multiple really cuts the voltage

> signal down quite substantially. I just got an MFB Dual ADSR and noticed

> that when i mult the Gate Output of my SH-101 by 3x, it drops the gate

> voltage down to where the MFB won't trigger anymore. Basically, i had it

> set up where the gate out went into the A-180, then the first 2 outs

> went to the two Gate Ins on the MFB and the 3rd out went to my A-140 EG.

> If i pulled the cable out that went to the A-140, the MFB would fire,

> but not with it inserted. However, it works if i use my 303, which has a

> hotter Gate signal. I checked the gate out voltages coming of the 101

> direct, then at each output of the A-180. I got the following:

>

> Direct Out from 101: 11.56v

> Into the A-180 w/no other cables inserted: same as above

> with 2 cables inserted: 7v

> with 3 cables inserted: 5.5v

>

> This is quite a bit of voltage drop if you ask me. Is there anyway to

> mod the A-180 so it doesn't eat up all my voltages?

>

> -Larry



Some basic information about this issue:



The A-180 is a passive module - nothing but 2x4 connected jack sockets. It

works as if one would simply connect the jack plugs manually of if one would

use an Y-cable. Consequently the reason for the behaviour described is

actually not the A-180 but the inputs and outputs of the modules. Ideally

the output of each module should be able to drive any load without any

losses (technically spoken: zero ohm output impedance or resistance). And

ideally the input of each module should not cause any load (technically

spoken: infinite ohm output impedance or resistance). But in reality the

situation is different: each output has a limited driving capability (e.g. 1

k output resistance) and each input causes a load (e.g. 100k for most VCO CV

inputs). As a rule of thumb the relation between input and output resistance

causes the voltage loss: if a 1k output drives a  100k input one has about

1/100 of voltage drop. Not much for a gate voltage but already too much for

a VCO control input(that's why special buffers like A-185-1/2 are

recommended for VCO control). If more than one input is connected to an

output (e.g. by means of an A-180) the resulting load becomes smaller: as a

rule of thumb the load is divided by two if both inputs have the same load.

Otherwise it's a bit more difficult to calculate. For example: the A-140 has

a 10k load at it's input. If two A-140 inputs are connected together the

resulting load is 5k. If the resistance of the output that drives the two

A-140 is also in the 5k range the output voltage is divided by two. If it

was 12V without load it becomes ~ 6V with two A-140.



In the A-100 we payed attention that (with some exceptions) each source is

able to drive a customary number of units (e.g. 3-4 A-140 from a gate

source). The output resistance of A-100 modules is typ. 1k and the input

resistance 10k or more (e.g. 100k for VCO control inputs). Consequently

within the A-100 no problems with gate signals should occur. For voltages

that drive VCOs buffer modules (A-185-1/2) are recommended. These have a

much lower output impedance. If an external voltage source is used the

situation may be different. One needs to know the output impedance to

calculate the expected voltage drop. To be on the safe side a buffer module

can be used (even for gate signals).



Best wishes

Dieter Doepfer





    
     

    
    


 



  






      

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